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On Sun, 08 Mar 2009 18:18:30 -0700, Gary G <see.signature@bottom
On Sun, 08 Mar 2009 17:45:25 -0700, Gary G <see.signature@bottomwrote:
Additional info.
In a vial of parallel sides, the contents are 2cm x 1.2cm.
These are pretty close measurements.
TIA.
Kiss French. Drink California.
gary at gaugler dot com
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On Mon, 9 Mar 2009 11:38:57 +0000, David Littlewood <...@nospam.demon.co.uk
In article <...@bottom.?.invalid
Do you know the density of the material? If so, then the total weight,
the volume of each sphere and the material density will allow you to
calculate how many spheres there are, and thus (from this number times
area of each sphere, 4*pi*r^2) the total surface area present.
Alternatively, if you don't know what the material is, and hence don't
know it's density, you can work it out from the volume data you quoted
in the second post.
David
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David Littlewood
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On Mon, 9 Mar 2009 17:06:53 +0000, David Littlewood <...@nospam.demon.co.uk
In article <...@bottom.?.invalidPer my previous post: "Alternatively, if you don't know what the
material is, and hence don't know it's density, you can work it out from
the volume data you quoted in the second post."
Sorry, must have been half asleep. Of course, if you do this, you will
have to apply a correction for the packing fraction, which IIRC from the
wiki page is 0.74 for identical spheres and approximately 0.64 for
randomly sized ones. You need to find the density of the solid material,
not the average density of the powder...
Hope this did not send you in a wrong direction.
David
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David Littlewood
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On Mon, 09 Mar 2009 18:22:48 -0700, Gary G <see.signature@bottom
On Mon, 9 Mar 2009 17:06:53 +0000, David Littlewood
<...@nospam.demon.co.uk
Thank you Richard and thank you David. I think this leads to a
solution. The diameter of the spheres are consistent within less than
a 0.5% variation. Of course I took LM and SEM pix of these spheres.
But they don't really play into the actual question.
Thanks again.
gg
Kiss French. Drink California.
gary at gaugler dot com
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